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free fall ellipse or parabola?
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free fall ellipse or parabola?
The Next CEO of Stack OverflowHohmann transfer orbit questionsPractical limits on the size of orbiting objects: could two pebbles orbit each otherCan a very small portion of an ellipse be a parabola?Are retrograde capture orbits “easier” than prograde capture orbits?How does acceleration affect the orbit?Prove that orbits are conic sectionsDo the planets really orbit the Sun?Meaning of the focus of an elliptical orbitHow did people get ellipses of Newton's equations of motion and gravitation?Is spacetime in an ellipse around a massive object, or does it just slope down towards the massive object?
$begingroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics gravity orbital-motion projectile free-fall
edited 35 mins ago
Aaron Stevens
13.7k42250
asked 55 mins ago
user56930user56930
174
$endgroup$
add a comment |
$begingroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics gravity orbital-motion projectile free-fall
edited 35 mins ago
Aaron Stevens
13.7k42250
asked 55 mins ago
user56930user56930
174
$endgroup$
add a comment |
$begingroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics gravity orbital-motion projectile free-fall
edited 35 mins ago
Aaron Stevens
13.7k42250
asked 55 mins ago
user56930user56930
174
$endgroup$
Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.
newtonian-mechanics gravity orbital-motion projectile free-fall
newtonian-mechanics gravity orbital-motion projectile free-fall
edited 35 mins ago
Aaron Stevens
13.7k42250
asked 55 mins ago
user56930user56930
174
edited 35 mins ago
Aaron Stevens
13.7k42250
asked 55 mins ago
user56930user56930
174
edited 35 mins ago
Aaron Stevens
13.7k42250
edited 35 mins ago
Aaron Stevens
13.7k42250
edited 35 mins ago
Aaron Stevens
13.7k42250
13.7k42250
asked 55 mins ago
user56930user56930
174
asked 55 mins ago
user56930user56930
174
asked 55 mins ago
user56930user56930
174
174
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
$endgroup$
add a comment |
$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
$endgroup$
add a comment |
$begingroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
$endgroup$
The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.
On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.
At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
answered 48 mins ago
Cort AmmonCort Ammon
24k34779
24k34779
add a comment |
add a comment |
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